//给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。 
//
// 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。 
//
// 此外，你可以假设该网格的四条边均被水包围。 
//
// 
//
// 示例 1： 
//
// 
//输入：grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
//]
//输出：1
// 
//
// 示例 2： 
//
// 
//输入：grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
//]
//输出：3
// 
//
// 
//
// 提示： 
//
// 
// m == grid.length 
// n == grid[i].length 
// 1 <= m, n <= 300 
// grid[i][j] 的值为 '0' 或 '1' 
// 
// Related Topics 深度优先搜索 广度优先搜索 并查集 
// 👍 1192 👎 0

/**
 * @author DaHuangXiao
 */
package leetcode.editor.cn;

public class NumberOfIslands {
    public static void main(String[] args) {
        Solution solution = new NumberOfIslands().new Solution();
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        private char[][] grid;
        public int numIslands(char[][] grid) {
            this.grid=grid;
            int count =0;
            for (int i=0;i<grid.length;i++){
                for (int j=0;j<grid[0].length;j++){
                    if (grid[i][j]=='1') {
                        dfs(i, j);
                        count++;
                    }
                }
            }
            return count;
        }
        public void dfs(int i, int j){
            if (i>=grid.length || i<0 || j>=grid[0].length || j<0 || grid[i][j]=='0'){
                return;
            }
            grid[i][j]='0';
            dfs(i+1,j);
            dfs(i-1,j);
            dfs(i,j+1);
            dfs(i,j-1);
        }
    }
//leetcode submit region end(Prohibit modification and deletion)

}